# A unit closed ball in a n.v.s is totally bounded?

Let $X$ be an infinite-dimensional normed linear space. $\overline{B}(0,1):=\{x\in X:||x||\leq 1\}$. My question is, $\overline{B}(0,1)$ is totally bounded?

If $X$ is complete, that is to say, $X$ is a Banach space, then $\overline{B}(0,1)$ is closed, so $\overline{B}(0,1)$ is complete. But $\overline{B}(0,1)$ is not compact, so $\overline{B}(0,1)$ is not totally bounded.

If $X$ is not complete, then $\overline{B}(0,1)$ is totally bounded?

def: $(X,d)$ is a metric space, $A$ is a subset of $X$, $A$ is called totally bounded, if for any $\epsilon>0$, there exist a finite subset $F\in X$, such that $A\subseteq\bigcup\limits_{a\in F} B(a,\epsilon)$

Thanks very much for any help.

The answer is no for every infinite-dimensional normed linear space $X$. One can show this by employing Riesz's Lemma. We construct a sequence $(x_n)_{n=1}^\infty$ in the closed unit ball $B$ of $X$ such that $$\|x_m - x_n\| > 1/2,$$ if $m \neq n$. Then no finite collection of balls having radius $1/4$ can cover $B$ (for they can't cover $\{x_n\}_{n=1}^\infty$).
• Suppose $B$ could be covered by a finite collection of balls having radius $1/4$. Then there would be positive integers $m\neq n$ such that $x_m$ and $x_n$ are in the same ball (pigeonhole principle). By the triangle inaquality we would have $\|x_m−x_n\| \leqq 1/2$, which can't happen. So the definition fails for $\epsilon=1/4$. It's not only that $(x_n)_{n=1}^\infty$ won't have a convergent subsequence, but each term is at distance $>1/2$ from each of the others. – user179514 May 7 '16 at 14:36